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Please give me step-by step solution. Answers are given, but I need detailed exp

ID: 1720879 • Letter: P

Question

Please give me step-by step solution. Answers are given, but I need detailed explanation...

"If something can go wrong, it will go wrong." This funny saying is called Murphy's law. Let's interpret this to mean "If something can go wrong, there is a very high probability that it will eventually go wrong."

Suppose we look at the event of having an automobile accident at some time during a day's commute. Let's assume that the probability of having an accident on a given day is 1 in a thousand or 0.001. That is, in your town, one of every thousand cars on a given day is involved in an accident (including little fender-benders). We also assume that having (or not having) an accident on a given day is independent of having (or not having) an accident on any other given day. Suppose you commute 43 weeks per year, 5 days a week, for a total of 215 days each year. In the following parts, write each probability in decimal form rounded to three places.

0.806

0.194



(c) Repeat part (a) for a 10-year period and for a 30-year period.

0.116

0.002


Repeat part (b) for a 10-year period and for a 30-year period.

0.884

0.998


(d) Does your work support the idea that there is a mathematical basis for Murphy's law as we interpreted it?

10-year period     (No Response)

0.116

30-year period     (No Response)

0.002

Explanation / Answer

a) Probability of having an accident = 1/1000 = 0.001 = P

Probability of not having an accident = 1 - P = 0.999

Probability that you have no accident over a year = (P(no accident))^(215) [ since the events are given independent] = (0.999)^(215) = 0.8064

b) Probability that you have atleast one accident = 1 - P(having no accident)

=> 1 - Previous answer

=> 1 - 0.8064

=> 0.1936

c) For 10 year period

Probability that you have no accident over a 10 year = (P(no accident))^(2150) [ since the events are given independent] = (0.999)^(2150) = 0.1163

Probability that you have atleast one accident in 10 year = 1 - previous answer = 1 - 0.1163 = 0.8836

For 30 year period

Probability that you have no accident over a 30 year = (P(no accident))^(6450) [ since the events are given independent] = (0.999)^(6450) = 0.0015

Probability that you have atleast one accident in 10 year = 1 - previous answer = 1 - 0.0015 = 0.9984

d) Yes, the probability of having an accident is very small, but as the time has beeen increased to 30 years, the probability of having the accident is approached to almost equal to 1